Midterm Exam III
CHEM 181: Introduction to Chemical Principles
November 25, 2014 Answer Key
1. For reference, here are the pKa values for three weak acids:
OH
HC
H
N
O
CH
HC
HC
C
CH
H 3C
CH
HC
OH
CH
N
O
O
pKa = 15
pKa = 4.8
pKa = 7.2
Now consider this list of compounds:
OH
H
C
O
H 3C
O
C
C
H2
HC
N
C
C
CH
O
HC
OH
H 2C
CH
C
CH2
Cl
O
OH
C
O
N
HC
C
OH
C
C
C
C
H2
C
H2
HC
CH
OH
HC
CH
C
H
1
(a) An unknown acid has a pKa of 2.5 and the following IR spectrum:
i. Identify this compound from the list on page 2 (redraw the Lewis
structure below).
O
N
C
C
C
H2
OH
This compound has a C≡N stretch in the IR spectrum and a pKa
significantly more acidic than the example carboxylic acid.
ii. Circle the most acidic proton of this compound.
iii. Draw any resonance structures that are important in determining the
acidity of this proton.
O
O
N
N
C
C
C
C
H2
O
C
C
H2
O
iv. Label any peaks on the IR spectrum that you can assign (there will
likely only be a few.)
It’s unclear whether the structure near 3000 cm−1 is the C–H stretch
or simply part of the O–H feature. The C≡N stretch is important.
2
(b) An unknown acid has a pKa of 9.5 and the following IR spectrum:
i. Identify this compound from the list on page 2 (redraw the Lewis
structure below).
H
N
O
C
O
C
H 2C
CH2
While there must be a C=O bond, the pKa is too high here (it is too
weak an acid) for the molecule to include a COOH.
ii. Circle the most acidic proton of this compound.
iii. Draw any resonance structures that are important in determining the
acidity of this proton.
O
N
C
H 2C
C
O
N
O
C
CH2
H 2C
C
CH2
O O
N
C
H 2C
C
O
CH2
The major resonance structures that move the negative charge onto
the oxygens are the reason this N–H bond is acidic (much more so
than the example compound.)
iv. Label any peaks on the IR spectrum that you can assign (there will
likely only be a few.)
The N–H stretch feature is too high in frequency (max at ∼3100–
3200 cm−1 , but spreading out between 3000 and 3500 cm−1 ) to be
3
an sp3 C–H stretch. There is a typo on the IR peak list that says
amines have a C–H stretch at this frequency, so “amine C–H stretch”
was accepted along with “N–H stretch” as an answer.
2. For each of the following, rank the compounds by most acidic (basic) to least
acidic (basic). Give a short explanation for each ranking.
(a) Write “strongest” next to the strongest acid and “weakest” next to the
weakest.
H 2C
O
O
C
C
C
H
OH
H 3C
O
H 2C
C
C
H
OH
H
weakest
strongest
The addition of a CH2 = slightly increases the acidity of the carboxylic
acid, explained either with a (very poor) additional resonance structure or
the difference between the sigma bond to an sp2 instead of an sp3 orbital.
The rightmost compound is the weakest because C–H hydrogens are almost
always less acidic than O–H hydrogens.
(b) Write “strongest” next to the strongest acid and “weakest” next to the
weakest.
Cl
O
O
C
C
C
H2
OH
H 3C
HC
CH
HC
OH
OH
HC
CH
weakest
strongest
The phenol (on the right) is the weakest because the charge remains mostly
on the O in the conjugate base, with only minor resonance structures
that locate the lone pair on carbon atoms. The carboxylic acids have two
major O− resonance structures, and the addition of the electronegative Cl
increases acidity.
4
(c) Write “strongest” next to the strongest BASE and “weakest” next to the
weakest BASE. (It is the NH2 group that acts as the base.)
O
O
O
HC
N
O
CH
C
C
C
HC
N
NH 2
CH
HC
CH
H 2C
C
HC
CH
NH 2
CH2
H 2C
CH
H 2C
NH 2
CH2
strongest
weakest
The strongest base will have the weakest conjugate acid; the weakest base,
the strongest acid. The conjugate acids are made by adding H+ to the
NH2 ; the question of whether the NH2 is itself acidic is unrelated.
The key here is that the NH+
3 aromatic compounds do not have minor resonance structures that double-bond the C and N, while the NH2
compounds—their conjugate bases—do:
(and 2 others)
NH 3
NH 2
vs.
NH 2
The NO2 group opposite the NH2 makes the unprotonated base even more
stable relative to the protonated acid. This gives the strongest acid, and
thus the weakest base.
O
N
NH 2
O
Resonance involving the NO2 is not possible with the molecule in the middle position, making the conjugate acid weaker and thus the base stronger.
No resonance at all happens with the rightmost compound, making the
protonated acid the weakest and thus the base the strongest. (Grading
was lenient on the relative strength of the third compound with respect to
the other two.)
5
3. Match the compounds on the following page to the 1 H NMR spectra on this
page. (All peaks are shown. Ignore very small spikes in the baseline, as well as
the calibration peak at 0 ppm.)
6
4. Acetic acid (CH3 COOH) is a weak acid, and pyridine (C5 H5 N) is a weak base.
Ka (CH3 COOH) = 1.8 × 10−5 M
Kb (C5 H5 N) = 1.7 × 10−9 M
If you start with 0.1 L of 0.1 M pyridine in water:
(a) How many moles of acetic acid do you have to add to convert 95% of the
pyridine to its conjugate acid form (C5 H5 NH+ )?
Acetic acid is the strongest acid in solution (much stronger than water)
and pyridine the strongest base (much stronger than water), so the reaction
that matters is
HAc + pyr Ac− + pyrH+
The K for this reaction is
K=
[Ac− ][pyrH+ ]
[HAc][pyr]
We can get the terms we need in the numerator and denominator like so
Ka (HAc)Kb (pyr) =
=
=
K =
=
=
[Ac− ][H+ ] [pyrH+ ][OH− ]
[HAc]
[pyr]
−
+
[Ac ][pyrH ] +
[H ][OH− ]
[HAc][pyr]
K · Kw
Ka (HAc)Kb (pyr)
Kw
1.8 × 10−5 · 1.7 × 10−9
1 × 10−14
3.06
The initial pyridine concentration is 0.1 M, so converting 95% of it means
[pyr] = 0.005 M
and
7
[pyrH+ ] = 0.095 M
These are equilibrium concentrations and let us solve for [Ac− ] and [HAc].
[Ac− ][pyrH+ ]
[HAc][pyr]
[Ac− ] [pyrH+ ]
[HAc] [pyr]
[Ac− ] 0.095
[HAc] 0.005
[Ac− ]
· 19
[HAc]
[Ac− ]
[HAc]
= 3.06
= 3.06
= 3.06
= 3.06
= 0.1611
Concept check:
i. If we added acetic acid to a strong base, all of the HAc will convert to
Ac− (at least until there is no strong base left; if only 95% has been
used up, there can be no HAc.)
ii. If we added acetic acid to a very weak base like water, then there would
only be a small amount of Ac− formed (because acetic acid is a weak
acid), with almost everything remaining as HAc.
iii. Since pyridine is a weak base, there is significant but not complete
conversion of HAc to Ac− .
Stoichiometry says that for every pyrH+ created, we must also create an
Ac− (the reaction with water is negligible), so
[Ac− ] = [pyrH+ ] = 0.095 M
At equilibrium,
[Ac− ]
= 0.1611
[HAc]
and so
[HAc] = 0.59 M
The total amount of acetic acid added is equal to the combined amount of
HAc and Ac− at equilibrium, so
[HAc]added = [HAc]eq + [Ac− ]eq = 0.685
and with a volume of 0.1 L
n(HAc) = 0.685 M × 0.1 L = 0.0685 mol
8
(b) What will the pH of the solution be at this point?
You don’t need part (a) to calculate the pH. Instead,
Kb (pyr) =
1.7 × 10−9 =
=
[OH ] =
−
[H+ ] =
=
pH =
[pyrH+ ][OH− ]
[pyr]
[pyrH+ ]
[OH− ]
[pyr]
19 · [OH− ]
8.95 × 10−11 M
Kw
[OH− ]
1.1 × 10−4 M
3.95
(This is essentially the same as using the Henderson-Hasselbalch equation.)
5. HCN is a gas that makes a weak acidic solution when dissolved in water.
Ka (HCN) = 6.2 × 10−10 M
The Henry’s law constant for the reaction
H2 O(`)
HCN(g) −→ HCN(aq)
is
kH = 8.3 M atm−1
A buffer solution with 0.1 M HCN(aq) is at a pH of 9.0. 100 mL of this solution
is placed in a sealed 4 L bottle.
What is the equilibrium pressure of HCN(g) in this bottle at 300 K? (Assume
the pH remains constant.)
The equilibria to consider are
[H+ ][CN− ]
= 6.2 × 10−10 M
[HCN]
and
[HCN(aq)]
= 8.3 M atm−1
PHCN
Because a pH of 9 is constant, [H+ ] = 1 × 10−9 M, which means
[1 × 10−9 M][CN− ]
= 6.2 × 10−10 M
[HCN]
9
and
[CN− ]
= 0.62
[HCN]
If all of the HCN remains in solution, then
[HCN] + [CN− ]
[CN− ]
[HCN]
[CN− ]
[HCN] + 0.62 [HCN]
[HCN]
= 0.1 M
= 0.62
= 0.62 [HCN]
= 0.1 M
= 0.062 M
Use this number to calculate the gas pressure:
[HCN(aq)]
= 8.3 M atm−1
PHCN
[HCN(aq)]
PHCN =
8.3 M atm−1
0.062 M
=
8.3 M atm−1
= 7.4 × 10−3 atm
Now ask the question: does producing this much gas decrease the amount of
HCN in solution?
n =
PV
RT
(7.4 × 10−3 atm)(3.9 L)
(0.08206 L atm mol−1 K−1 )(300 K)
= 1.2 × 10−3 mol
=
With 100 mL of 0.1 M HCN, we only had 1 × 10−2 mol to start with; after
producing this much gas, only 8.8 × 10−3 mol (100 mL at 0.088 M) remains in
solution. The [HCN(aq)] is reduced proportionally:
[HCN] = 0.88 · 0.062 = 0.055 M
This, in turn, decreases the pressure accordingly:
PHCN = 0.88 · 7.4 × 10−3 = 6.5 × 10−3 atm
10
(You can iterate more times, though it is not necessary: the reduced pressure
means that only 1.03 × 10−3 mol is removed from solution, thus a more accurate
concentration for HCN is 0.056 M and pressure is 6.6 × 10−3 atm.)
Alternately, using more algebra avoids iteration. Solve for number of moles of
gas as a function of [HCN(aq)]:
[HCN(aq)]
8.3 M atm−1
PV
=
RT
PHCN =
ngas
(3.9 L)
(0.08206 L atm mol−1 K−1 )(300 K)
[HCN(aq)]
=
· 0.1574 mol atm−1
8.3 M atm−1
= 0.019 L · [HCN(aq)]
= P·
The total number of moles is divided between gas, aqueous HCN, and aqueous
CN− :
ntot = ngas + nHCN(aq) + nCN−
0.01 = 0.019 · [HCN(aq)] + 0.1 · [HCN(aq)] + 0.1 · [CN− (aq)]
= 0.019 · [HCN(aq)] + 0.1 · [HCN(aq)] + 0.062 · [HCN(aq)]
0.01
[HCN(aq)] =
0.019 + 0.1 + 0.062
= 0.0552 M
and
PHCN =
[HCN(aq)]
= 6.65 × 10−3 atm
8.3 M atm−1
11